﻿ Verify the following identity \dfrac{\sin 4x+\sin 6x}{\sin 4x-\sin 6x} =-\tan 5x\cot x

#### Question

Verify the following identity \dfrac{\sin 4x+\sin 6x}{\sin 4x-\sin 6x} =-\tan 5x\cot x

Collected in the board: Trigonometry

Steven Zheng posted 2 years ago

Apply the sum-to-product identity for sin\alpha +\sin\beta to the numerator

\sin \alpha + \sin \beta = 2\sin\dfrac{\alpha +\beta }{2} \cos\dfrac{\alpha -\beta }{2}

Then,

\sin 4x+\sin 6x =2\sin\dfrac{4x+6x}{2}\cos\dfrac{4x-6x}{2}

\sin 4x+\sin 6x =2\sin 5x\cos(-x)

\because \cos(-x) =\cos x

\therefore \sin 4x+\sin 6x =2\sin 5x\cos x
(1)

Apply the sum-to-product identity for sin\alpha -\sin\beta to the denominator

\sin \alpha - \sin \beta = 2\cos\dfrac{\alpha +\beta }{2} \sin\dfrac{\alpha -\beta }{2}

Then,

\sin 4x-\sin 6x =2\cos\dfrac{4x+6x}{2}\sin\dfrac{4x-6x}{2}

\sin 4x-\sin 6x = 2\cos 5x\sin(-x)

\because \sin(-x)=-\sin x

\therefore \sin 4x-\sin 6x = -2\cos 5x\sin x
(2)

Substitute (1) and (2) into the LHS of the identity

\dfrac{\sin 4x+\sin 6x}{\sin 4x-\sin 6x}

=\dfrac{2\sin 5x\cos x}{-2\cos 5x\sin x}

=-\tan 5x\cot x

Now we have verified the identity

\dfrac{\sin 4x+\sin 6x}{\sin 4x-\sin 6x} =-\tan 5x\cot x

Steven Zheng posted 2 years ago

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